# Thread: Help of Intermediate Value Theorem

1. ## Help of Intermediate Value Theorem

Use the Intermediate Value Theorem to show that $x^5-x^2+1=0$has a solution.

Given:
$f(-1)=-1$ and $f(0)=1$

I understand what the IVT is and how to find a approximate solution using trial and error, but what exactly does it mean by "show?" Are they asking for a proof? I understand the concept of IVT but I can't translate my potential way of solving this into math symbols and such

Thanks

2. Originally Posted by c_323_h
Use the Intermediate Value Theorem to show that $x^5-x^2+1=0$has a solution.

Given:
$f(-1)=-1$ and $f(0)=1$
The function,
$f(x)=x^5-x^2+1$ is countinous on $[-1,0]$
By IVT,
For any $-1=f(-1)\leq k\leq f(0)=1$ there exists an $1 \leq x \leq 0$ such as, $f(x)=k$
Let $k=0$ and the proof is finished.

3. Originally Posted by ThePerfectHacker
The function,
$f(x)=x^5-x^2+1$ is countinous on $[-1,0]$
By IVT,
For any $-1=f(-1)\leq k\leq f(0)=1$ there exists an $1 \leq x \leq 0$ such as, $f(x)=k$
Let $k=0$ and the proof is finished.
This is what I tried. Let $f(x)=x^5-x^2+1=0$ . Then, by the IVT, there exists a number $c$ such that
$f(-1). Since the function is continuous over the interval $(-1,1)$ there exists a number $N$ such that $-1 and $f(c)=N$

I don't know if that proves anything though. I really don't have a direction when I try to prove things. I just copied the definition of IVT out of the book.

4. Let me try this again.

If $f(x)$ is a real function countinous on closed interval $[a,b]$, and let $k$ be any number between $f(a)$ and $f(b)$ then there must exist an $x$ on $[a,b]$ such as, $f(x)=k$.

Root location theorem: If $f(x)$ is countinous on $[a,b]$ and $f(a)f(b)<0$ (in other words opposite signs) then there is a solution to the equation $f(x)=0$ on $[a,b]$

Proof: By the condition we have that $f(a)$ and $f(b)$ have opposite signs. Thus, zero is a number between them. Thus, by the intermediate value theorem, $f(x)=0$ for some $x$ on $[a,b]$
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Returning to your problem.
1)Given $f(x)=x^5-x^2+x+1$ on $[-1,0]$
2)The function is countinous cuz all polynomials are.
3)And, $f(-1)=-1$ and $f(0)=1$
4)And, $f(-1)f(0)<0$
5)Condition for Root location theorem are satisfied.
6)Thus there exists a solution to $f(x)=0$.